confused about a enthalpy question

Calculate the  standard  enthalpy  change  of formation,  ∆fHө, of hydrogen sulfide using
the enthalpy change for Reaction 1, and the standard enthalpy changes of combustion
below.
Substance ∆cH ө / kJ mol−1
S(s) −296.8
H2(g) −285.8
2H2S(g)  +  3O2(g) 2SO2(g) + 2H2O(l)      ∆rH = –1125 kJ mol−1
I drew a hess cycle graph and i put my arrow inwards as I thought that even if we want to work out a value of formation our data is combustions so the arrows must point down \.
What i did was form a equation:
2x + (296x2) + (285.8x2) = -1125
2x = -2290.2
x = -1145.1 but this is wrong and the answer is -20.

Use definition of enthalpy change of combustion: 1 mole of _ react completely with oxygen.
So write the equation S(s) −296.8 and H2(g) -285.8 as:
S + O2 -> SO2 (-296.8)
H2 + 0.5O2 -> H2O (-285.8)
With:
2H2S(g)  +  3O2(g) -> 2SO2(g) + 2H2O(l)  (enthalpy change of reaction –1125)
We can half this, to become -562.5 to fit the enthalpy change of combustion definition. (As we are given combustion data).
H2S(g)  +  1.5O2(g) -> SO2(g) + H2O(l)  (enthalpy change of combustion –562.5)
Now if we combine: S + O2 -> SO2 (-296.8) with H2 + 0.5O2 -> H2O (-285.8)
S + 1.5O2 + H2 -> SO2 + H2O
-286.8 + -285.8 = -582.6
To find: H2 + S -> H2S
-582.6 - -562.5 = -20KJ mol-1
To simplify: for combustion
Do sum of reactant - sum of product.
From what you wrote, you just needed to make your (296x2) + (285.8x2) negative. If you did to that, rearranging for x would give -20.